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Potential divider and its Thevenin's circuit.

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(a) A potential divider is to supply a voltage of 10 V, constant to better than 10%, into a load which can draw current up to a maximum of 50 mA. Design a potential divider for a 40 V battery (internal resistance 1 ohm) that would guarantee these conditions.
(b) Determine Thevenin's circuit for the circuit of (a)

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(a) A potential divider is to supply a voltage of 10 V, constant to better than 10%, into a load which can draw current up to a maximum of 50mA. Design a potential divider for a 40 V battery (internal resistance 1 ohm) that would guarantee these conditions.
(b) Determine Thevenin's circuit for the circuit of (a)

(a) The potential divider supply circuit is shown in the diagram. For any load RL connected parallel to R2 the current and hence the voltage across R2 will decrease. For very high resistances the decrease in the voltage across R2 (supply voltage) will be very small.
When no load is connected, (RL = infinite) R1, R2, and r are in series and hence equivalent resistance of the circuit will be
R = r + R1 + R2

And the current in the circuit will be

I = ---- (1)

As the supply V2 is across R2 and it is to be 10 V we have

V2 = I*R2
Or 10 =
Gives 3R2 = R1 + r --- (2)

As the load is RL is connected as in diagram, the resistance of parallel combination is reduced and hence the potential drop at this combination (supply voltage) decreases. The maximum permissible decrease 10% of supply voltage of 10V hence the minimum supply voltage will be

Vmin = 10 - 10*(10/100) = 9 V

For this value of voltage the current in the load is to be 50 mA. Hence the load resistance will have a minimum value given by

Vmin = RL*IL

Gives RL = Vmin/IL = 9/(50*10-3) = 180 

And in this situation the current in R2 will be

I2 = 9/R2

Hence the total current in the circuit through the battery with minimum load will be

I' = I2 + IL = (9/R2) + 50*10-3

Now with this load resistance applying loop rule to the loop with V, R1, R2, and r we have

V = I'R1 + I2R2 + I'r

Or 40 = [(9/R2) + 50*10-3]*R1 + 9*1 + [(9/R2) + 50*10-3]*r

Or 31 = [(9/R2) + 50*10-3]*(R1 + 1)

Substituting value of R1 +1 from equation (2) we have

31 = [(9/R2) + 50*10-3]*3R2

Or 31 = 27 + 150*10-3*R2

Gives R2 = 4000/150 = 80/3 = 26.667 

And R1 =3R2 -1 = 79 

(b) Thevenine's circuit:

According to Thevenine's theorem, any complex circuit between two terminals can be represented by a voltage source and a resistance. The voltage between the two points when no external resistance is attached is called Thevenine's voltage. The Thevenine's resistance is the resistance between the terminals without considering the voltage in between.

Hence the Thevenine's voltage is the voltage between the points A and B when RL is not connected.

The current in the circuit will be

I = V/(R1+R2+r) = 40/(79+26.667+1)

Or I = 0.375 A

And hence the voltage across A and B will be

VTh = R2*I = 26.667*0.375 = 10V

This is the Thevenine's voltage

Thevenine's resistance for the circuit is the resistance of the circuit when A and B are short circuited hence

RTh = R1 + r = 79 +1 = 80 

Hence Thevenine's equivalent circuit will be as shown in the diagram.

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